Problem: What is the value of $\dfrac{d}{dx}\tan(x)$ at $x=\dfrac{\pi}{3}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{3}$ (Choice B) B $4$ (Choice C) C $\dfrac{\sqrt{3}}{2}$ (Choice D) D $2$
Let's first find $\dfrac{d}{dx}\tan(x)$. Then, we can evaluate it at $x=\dfrac{\pi}{3}$. Recall that the derivative of $\tan(x)$ is $\dfrac{1}{\cos^2(x)}$, or $\sec^2(x)$. Put another way, $\dfrac{d}{dx}[\tan(x)]=\dfrac{1}{\cos^2(x)}=\sec^2(x)$. [Is there a way to know this without memorizing?] Now let's plug in $x={\dfrac{\pi}{3}}$ : $\begin{aligned} &\phantom{=}\dfrac{1}{\cos^2\left({\dfrac{\pi}{3}}\right)} \\\\ &=\dfrac{1}{\left(\dfrac12\right)^2} \\\\ &=\dfrac{1}{\left(\dfrac14\right)} \\\\ &=4 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\tan(x)$ at $x=\dfrac{\pi}{3}$ is $4$.